Problem_Set_5_Solutions

Problem_Set_5_Solutions - 8. We use Eq. 7-12 for Wg and Eq....

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(c) With y = h = 5 R , at P we find 22 5 5(3.20 10 kg)(9.80 m/s )(0.12 m) 0.19 J Um g R ==× = . (d) With y = R , at Q we have (3.20 10 kg)(9.80 m/s )(0.12 m) 0.038 J g R = (e) With y = 2 R , at the top of the loop, we find 2 2(3.20 10 kg)(9.80 m/s )(0.12 m) 0.075 J g R = (f) The new information () v i 0 is not involved in any of the preceding computations; the above results are unchanged. 8. We use Eq. 7-12 for W g and Eq. 8-9 for U . (a) The displacement between the initial point and Q has a vertical component of h R downward (same direction as G F g ), so (with h = 5 R ) we obtain 4 4(3.20 10 kg)(9.80 m/s )(0.12 m) 0.15 J gg WF d m g R =⋅ = = × = G G . (b) The displacement between the initial point and the top of the loop has a vertical component of h – 2 R downward (same direction as G F g ), so (with h = 5 R ) we obtain 3 3(3.20 10 kg)(9.80 m/s )(0.12 m) 0.11 J dm g R = = × = G G .
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18. We place the reference position for evaluating gravitational potential energy at the relaxed position of the spring. We use x for the spring's compression, measured positively downwards (so x > 0 means it is compressed). (a) With x = 0.190 m, Eq. 7-26 gives 2 1 7.22 J 7.2 J 2 s Wk x =− ≈− for the work done by the spring force. Using Newton's third law, we see that the work done on the spring is 7.2 J. (b) As noted above, W s = –7.2 J. (c) Energy conservation leads to KU K U mgh mgx kx ii f f += + + 0 2 1 2 which (with m = 0.70 kg) yields h 0 = 0.86 m. (d) With a new value for the height == hh 00 21 7 2 .m , we solve for a new value of x using the quadratic formula (taking its positive root so that x > 0). mgh mgx kx x mg mg mgkh k ′=− + ¡ = ++ 0 2 2 0 1 2 2 b g which yields x = 0.26 m.
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34. The distance the marble travels is determined by its initial speed (and the methods of Chapter 4), and the initial speed is determined (using energy conservation) by the original compression of the spring. We denote h as the height of the table, and x as the horizontal distance to the point where the marble lands. Then x = v 0 t and hg t = 1 2 2 (since the vertical component of the marble's “launch velocity” is zero). From these we find xv h g = 0 2 . We note from this that the distance to the landing point is directly proportional to the initial speed. We denote v 0 1
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_5_Solutions - 8. We use Eq. 7-12 for Wg and Eq....

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